Lim as x approaches 0: (sin3x)/(2x-Sinx)
Taking the derivative of the numerator and denominator separately:
Numerator: d/dx(sin3x) = 3cos3x
Denominator: d/dx(2x - sinx) = 2 - cosx
Now, evaluate the limit using L'Hôpital's Rule:
Lim as x approaches 0: (3cos3x)/(2 - cosx)
Plugging in x = 0:
Lim as x approaches 0: (3cos(0))/(2 - cos(0))
= 3/2
Therefore, the limit as x approaches 0 of (sin3x)/(2x-Sinx) is 3/2.
Lim as x approaches infinity: x^-1 lnx
Taking the derivative of the numerator and denominator separately:
Numerator: d/dx(x^-1 lnx) = (1/x)lnx
Denominator: d/dx(1) = 0
Since the denominator is 0, we cannot apply L'Hôpital's Rule. However, we can still evaluate the limit:
Lim as x approaches infinity: x^-1 lnx
As x approaches infinity, the natural logarithm (lnx) grows without bound, so the overall limit is 0.
Therefore, the limit as x approaches infinity of x^-1 lnx is 0.
Lim x approaches infinity: x/ e^x
Taking the derivative of the numerator and denominator separately:
Numerator: d/dx(x) = 1
Denominator: d/dx(e^x) = e^x
Now, evaluate the limit using L'Hôpital's Rule:
Lim as x approaches infinity: 1/ e^x
As x approaches infinity, the exponential function e^x grows without bound, so the overall limit is 0.
Therefore, the limit as x approaches infinity of x/ e^x is 0.